Question: Let $f(x, y, z) = z\cos(y) + z^2x$ and $g(t) = (t, -t^2, -t)$. $h(t) = f(g(t))$ $h'(t) = $
Explanation: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(t) = \nabla f(g(t)) \cdot g'(t)$. $\begin{aligned} &g(t) = (t, -t^2, -t) \\ \\ &g'(t) = (1, -2t, -1) \\ \\ &\nabla f = (z^2, -z\sin(y), \cos(y) + 2zx) \\ \\ &\nabla f(g(t)) = \left( t^2, t\sin(-t^2), \cos(-t^2) - 2t^2 \right) \end{aligned}$ Substituting: $\begin{aligned} h'(t) &= t^2 - 2t^2\sin(-t^2) -\cos(-t^2) + 2t^2 \\ \\ &= 3t^2 + 2t^2\sin(t^2) - \cos(t^2) \end{aligned}$ Answer $h'(t) = 3t^2 + 2t^2\sin(t^2) - \cos(t^2)$